A projectile is shot at an angle of #pi/12 # and a velocity of #4 m/s#. How far away will the projectile land?

1 Answer
Mar 15, 2016

Answer is:

#s=0.8m#

Explanation:

Let the gravity acceleration be #g=10m/s^2#

The time traveled will be equal to the time it reaches its maximum height #t_1# plus the time it hits the ground #t_2#. These two times can be calculated from its vertical motion:

The initial vertical speed is:

#u_y=u_0sinθ=4*sin(π/12)#
#u_y=1.035m/s#

Time to maximum height #t_1#

As the object decelerates:

#u=u_y-g*t_1#

Since the object finally stops #u=0#

#0=1.035-10t_1#

#t_1=1.035/10#

#t_1=0.1035s#

Time to hit the ground #t_2#

The height during the rising time was:

#h=u_y*t_1-1/2*g*t_1^2#

#h=1.035*0.1035-1/2*10*0.1035^2#

#h=0.05359m#

The same height applies to the dropping time, but with the free fall formula:

#h=1/2*g*t_2^2#

#t_2=sqrt((2h)/g)#

#t_2=0.1035s#

(Note: #t_1=t_2# because of energy preservation law.)

The total time traveled is:

#t_t=t_1+t_2#

#t_t=0.1035+0.1035#

#t_t=0.207s#

The distance traveled in the horizontal plane has a constant speed equal to:

#u_x=u_0cosθ=4*cos(π/12)#
#u_x=3.864m/s#

Finally, the distance is given:

#u_x=s/t#

#s=u_x*t#

#s=3.864*0.207#

#s=0.8m#

P.S. For future problems identical to this one but with different numbers, you can use the formula:

#s=u_0^2*sin(2θ)/g#

Proof : we are basically going to use the same method inversely, but without substituting the numbers:

#s=u_x*t_t#

#s=u_0cosθ*2t#

#s=u_0cosθ*2u_y/g#

#s=u_0cosθ*2(u_0sinθ)/g#

#s=u_0^2*(2sinθcosθ)*1/g#

#s=u_0^2*sin(2θ)*1/g#

#s=u_0^2*sin(2θ)/g#